Question

The angle of depresion of the top and bottom of 50 m high building from the top of a tower ere 45°
and 60° respectively, Find the height of the tower and the horizontal distance between the tower and the
building (Use √3 - 1.73)​

Answer 1

Solution: tan ∅ = Perpendicular/Base

→ tan 45° = AE/ED

→ 1 = AE/ED

→ AE = ED

→ tan 60° = AB/BC

→ √3 = (AE + BE)/BC

Now, AE = DE and BE = DC (opposite sides of rect. BCDE)

→ √3 = (DE + DC)/BC

Now DE = BC (opposite sides of rect. BCDE)

→ √3 = (BC + DC)/BC

→ √3 = 1 + DC/BC

→ √3 = 1 + 50/BC

→ (√3 - 1) = 50/BC

→ BC = 50(√3 + 1)/(√3² - 1²)

→ BC = 25(√3 + 1)

→ BC = 68.25 m

Now, BC = ED = AE

→ AE = 68.25 m

→ AB = AE + BE

→ AB = 68.25 m + 50 m

→ AB = 118.25 m

Answer: Height of tower is 118.25 m and horizontal distance between building and towet is 68.25 m.

Answer 2

$\huge\tt{AnsweR}$

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Tan = 54° AE / ED

= AE / ED1

= AE / EDAE = ED

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Tan = 60° = AB / BC

= √3 = ( AE + BE ) / BCAE = DE , DE

= DE ( BCDE )

= √3 = ( DE + DC / DC )

= BE = BC ( BCDE )

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= √3 = ( 3 C + DC ) / BC

= √3 = 1 + DC / BC

= √3 = 1 + 50 / BC

= √3 - 1 = 50 / BC

= BC = 50 ( √3 + 1 ) / √3² - 1²

= BC = 25 ( √3 + 1 )

= BC = 68.25 m

= BC = ED = DE

= AE = 68.25 m

= AB = AE + BE

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= AB = 68.25 m + 50 m

= AB = 118.25 m

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